# What is cos x if cscx=4squareroot3/3 and x is in the interval (pi;3pi/2)?

### 3 Answers | Add Yours

Recall that sin(x) = 1/csc(x).

sin(x) = 1/(4sqrt(3)/3) = 3sqrt(3)/[4sqrt(3)*sqrt(3)] = 3sqrt(3)/(4)(3) = 3sqrt(3)/12 = sqrt(3)/4

Now lets visualize this in a right triangle. The sine is opposite over hypotenuse. So we can say that the opposite is sqrt(3) and the hypotenuse is 4.

Using the Pythagorean Theorem, we can solve for the adjacent side. We need it for the cosine, which is adjacent over hypotenuse.

x^2 + sqrt(3)^2 = 4^2

x^2 + 3 = 16

x^2 = 13

x = sqrt(13)

cos(x) would be adjacent over hypotenuse, so we have:

cos(x) = sqrt(13)/4 but in Quadrant 3, the interval (pi, 3pi/2), the cosine is negative.

**Therefore, the answer is: -sqrt(13)/4**

We have csc x = 4*sqrt 3/3 = 4/sqrt 3

sin x = 1/csc x = sqrt 3/4

cos x = sqrt(1 - (sin x)^2)

=> sqrt(1 - (sqrt 3/4)^2)

=> sqrt(1 - 3/16)

=> -sqrt (13/16)

We do not consider the positive value as cos x is negative in the interval (pi, 3pi/2)

**The value of cos x = (-sqrt 13)/4**

We'll recall the identity that gives the cosecant function:

csc x = 1/sin x => sin x = 1/csc x

Since the value of the sine function is negative in the 3rd quadrant, I presume that you've missed the minus sign when you've provided the value of the cosecant function.

sin x = -1/(4sqrt3/3)

sin x = -3/4*sqrt3

sin x = -3sqrt3/12

Now, we'll use the Pythagorean theorem to determine cos x:

cos x = - sqrt [1 - (sin x)^2]

cos x = -sqrt (1 - 27/144)

cos x = - sqrt(144-27)/12

cos x = - 3sqrt13/12

**The requested value of he cosine function is negative, since x is an angle that belongs to the 3rd quadrant, and it is: cos x = - 3sqrt13/12.**